I'm not 100% sure on this, but I would go with C) NaCl.
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.

Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.

Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.

Answer:
Explanation:
1.)azeotrope is a mixture of two or more liquid components under constant boiling, it has a constant mole fraction composition of present component which can be homogeneous or heterogeneous.
2.)the condition which it's best performed when there's liquids that is non-volatile which boils higher than other liquids with at least 26 degrees .
steam azentropic distillation
3.During a steam distillation, How to know if the organic compound is still coming over is when you see the solution becoming cloudy or when there is existence of two layers.
4.)The end of the steam distillation, the receiving flask should contain two layers of liquid, and the chemical identity of these two liquids most contain
A.) Layers that are mostly water H2O
B.) Layers that are mostly products
5.)What is the purpose of adding 10% sodium carbonate solution to the distillate if it is acidic to litmus is to neutralize the distillate.
Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L